Problem K: Yikes -- Bikes!】的更多相关文章

http://acm.upc.edu.cn/problem.php?id=2780 昨天做的题,没过……!!!伤心……题意:给你n个单位,n-1组关系,让你单位换算……解题思路:Floyd算法自己听别人说用Floyd算法,然后自己默默的用有向图写……但是!!!Floyd算法不能用有向图……!所以只能在其相反的转化中标记为负的,在进行时特殊处理一下,最后便利找出能进行单位转化的那组单位,然后进行大小排序,最后就莫名其妙的哦过了……!!! #include <iostream> #include…
Problem K. Kitchen Robot Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100610 Description Robots are becoming more and more popular. They are used nowadays not only in manufacturing plants, but also at home. One programmer wit…
题目链接:https://codeforces.com/contest/1089/problem/K time limit per test: 2 seconds memory limit per test: 512 megabytes King Kog got annoyed of the usual laxity of his knights — they can break into his hall without prior notice! Thus, the King decided…
题目链接:http://codeforces.com/gym/101981/problem/K Your friend has made a computer video game called “Kangaroo Puzzle” and wants you to give it a try for him. As the name of this game indicates, there are some (at least 2) kangaroos stranded in a puzzle…
Problem K. UTF-8 Decoder 题目连接: http://opentrains.snarknews.info/~ejudge/team.cgi?SID=c75360ed7f2c7022&all_runs=1&action=140 Description UTF-8 is a character encoding capable of encoding all possible characters, or code points, in Unicode. Nowadays…
Problem K. KMC Attacks 题目连接: http://codeforces.com/gym/100714 Description Warrant VI is a remote planet located in the Koprulu Sector. Warrant VI features a strange huge field, which was discovered by the Kel-Morian Combine (KMC). The KMC is involved…
题目:Problem K. PiecemakingInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 512 mebibytesThe civil war in Berland continues for five years already. The United Nation decided to end the bloodshed.Berland consists o…
任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6342 Problem K. Expression in Memories Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2150    Accepted Submission(s): 772Special Judge Problem De…
6342.Problem K. Expression in Memories 这个题就是把?变成其他的使得多项式成立并且没有前导零 官方题解: 没意思,好想咸鱼,直接贴一篇别人的博客,写的很好,比我的垃圾好多了... HDU 6342(模拟) 贴一下一个队友的代码: //1011-6342-模拟 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include…
Problem K: 负2进制 Time Limit: 2 Sec Memory Limit: 128 MB Submit: 51 Solved: 6 [Submit][Status][Web Board] Description 如果我16岁,我可以悄悄的说我好喜欢你:如果我26岁,我可以大声告诉你我很爱你:可惜我6岁,我什么都给不了你,我还要上小学. 我们都知道2进制,每一位的权值如下: 1 2 4 8 16 32 64 现在我们定义一种-2进制,每一位的权值如下: 1 -2 4 -8 16…
Problem K: 搜索基础之棋盘问题 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 92  Solved: 53[Submit][Status][Web Board] Description 在一个给定形状的棋盘(形状可能是不规则的)上面摆放棋子,棋子没有区别.要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列,请编程求解对于给定形状和大小的棋盘,摆放k个棋子的所有可行的摆放方案C. Input 输入含有多组测试数据.每组数据的第一行是两…
Problem K: Deadline Time Limit: 2 Sec Memory Limit: 1280 MB Submit: 1106 Solved: 117 [Submit][Status][Web Board] Description There are N bugs to be repaired and some engineers whose abilities are roughly equal. And an engineer can repair a bug per da…
Problem K — limit 1 second Tournament Wins 这个题就是有2^n队伍,他现在的实力水平是第k位,采用的是淘汰制 问一下你他的胜场数的期望 这人能 win>=i 场的概率就是和它同一个半区的 2^i 个人都比他弱啊 所以去枚举这个2^i,E(X)=\sum P(X>=i) 直接搞一下就好了 我每次的概率都在前一次的算,这样避免了多次枚举,之后求下差就好,所以这个算法是2^n的 #include <bits/stdc++.h> using nam…
Problem K. Expression in Memories Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0Special Judge Problem Description Kazari remembered that she had an expression s…
K - Sliding Window Time Limit: 18000/6000MS (Java/Others)     Memory Limit: 131072/131072KB (Java/Others) Submit Status An array of size n≤106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the…
Problem Description The local toy store sells small fingerpainting kits with between three and twelve 50ml bottles of paint, each a different color. The paints are bright and fun to work with, and have the useful property that if you mix X ml each of…
K - 王之盛宴 Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others) Submit Status 王征战回来了,为了庆祝胜利,王准备请大家吃饭! 于是n个人来到了一家豪华餐厅,这家餐厅有一张长————————长的桌子,每个人只能坐在桌子的南北两侧 一行人中,有p对A关系,m对B关系,如果u和v有A关系,则u和v必须坐在不同侧,如果u和v有B关系,则u和v必须坐在同侧 如果一种座位安…
K - 摩天轮 Time Limit: 10000/4000MS (Java/Others)     Memory Limit: 262143/262143KB (Java/Others) Submit Status 一天,冬马被春希和雪菜拉着去一起去游乐园玩. 经过了各种过山车的洗礼后,三人决定去坐摩天轮休息下. 这是一个巨大的摩天轮,每一个车厢能坐任意多的人.现在,等着坐摩天轮的有n个人(包含他们3人),摩天轮还有m个车厢可以坐人.每个人都有自己肥胖程度,出于某些原因,胖子和瘦子坐在同一节车…
K - 秋实大哥の恋爱物语 Time Limit: 5000/2000MS (Java/Others)     Memory Limit: 32000/32000KB (Java/Others) Submit Status 传说有这么一个故事! 在一个月白风清的晚上,秋实大哥约一位他心仪的妹子一起逛校园,浪漫的秋实大哥决定在当晚对妹子表白.“XXXXX...”,秋实大哥温情地说完了准备已久的话.而妹子决定用一种浪漫的方式接受秋实大哥(其实妹子早已对秋实大哥动心,这一刻她早已迫不及待了,但还是决定…
之前几乎没写过什么这种几何的计算题.在众多大佬的博客下终于记起来了当时的公式.嘚赶快补计算几何和概率论的坑了... 这题的要求,在对两圆相交的板子略做修改后,很容易实现.这里直接给出代码.重点的部分有:两圆在相离(或外交)时输出第一个圆的面积.内涵(或内切)则需要分类讨论,是羊的圈大.还是狼的圈大.以下是代码: #include<iostream> #include<cmath> #include<stdio.h> using namespace std; int ma…
题意:让你构造一个n个点的简单多边形,使得所有点是整点,并且所有边长是整数,并且没有边平行于坐标轴. 就利用勾股数,如下图这样构造即可,n为偶数时,只需矩形拼成,n为奇数时,封上虚线边即可. #include<cstdio> using namespace std; struct Point{ int x,y; Point(const int &x,const int &y){ this->x=x; this->y=y; } Point(){} }p[1005];…
发现由于角的度数和边的长度有限制,那俩圆如果放得下的话,必然是塞在两个角里. 于是预处理n个圆心的位置(注意要判断那个圆会不会和其他的边界相交),然后n^2枚举俩角即可. #include<cstdio> #include<cmath> using namespace std; #define EPS 0.00000001 struct Point{ double x,y; double length(){ return sqrt(x*x+y*y); } }a[2010]; typ…
#include<stdio.h> int main() { int n,i,j,k,t,x,q,p; while(scanf("%d",&n)!=EOF) ;i<=n;i++) { ;j<=n-i;j++) printf(" "); ;k<=*i-;k++) printf("%d",i); printf("\n"); } ;x<=*n-;x++) { ;p<=x-n;p++)…
目录 @description@ @solution@ @accepted code@ @details@ @description@ 给定 n 个点,第 i 个点位于 (xi, yi). 在第 i 个点与第 j 个点之间建边费用为 xi*xj + yi*yj. 求最小生成树. Input 第一行一个整数 T (1≤T≤2000),表示数据组数. 每组数据给定一个整数 n(2≤n≤100000),表示点数.保证 ∑n≤10^6. 接下来 n 行,每行两个整数 xi, yi(1≤xi,yi≤10^…
题目链接:https://codeforces.ml/gym/102428/attachments/download/9820/statements-en.pdf 题意:构造一个多项式使得外星人编号的函数值小于零,人类的编号函数值大于零. 思路:零点存在问题,我们只要找出字符串中有多少A-H或者H-A的变化就是零点个数,然后利用求多项式板子的代码即可. 1 #include<iostream> 2 #include<algorithm> 3 #include<vector&g…
Description A math instructor is too lazy to grade a question in the exam papers in which students are supposed to produce a complicated formula for the question asked. Students may write correct answers in different forms which makes grading very ha…
用前序中序建立二叉树并以层序遍历和后序遍历输出 writer:pprp 实现过程主要是通过递归,进行分解得到结果 代码如下: #include <iostream> #include <queue> #include <cstdio> #include <cstring> using namespace std; const int N = 1000; struct tree { tree* l; tree* r; int data; tree() { l…
#include<stdio.h> int main() { int n; ]; while(scanf("%d%*c",&n)!=EOF) { while(n--) { ,num2=,num3=,num4=,num5=; gets(a); ;a[i]!='\0';i++) { switch(a[i]) //在这用switch语句会比较方便 { case 'a': num1++;break; case 'e': num2++;break; case 'i': num…
Description There is a robot, its task is to bury treasures in order on a N × M grids map, and each treasure can be represented by its weight, which is an integer. The robot begins to bury the treasures from the top-left grid. Because it is stupid, i…
题意:有\(n\)堆物品,每次可以将两堆捆成一堆,新堆长度等于两个之和,每次消耗两个堆长度之和的长度,求最小消耗使所有物品捆成一堆. 题解:贪心的话,每次选两个长度最小的来捆,这样的消耗一定是最小的,但是我们需要一个容器来存这些数,这时候很明显要用到优先队列(小根堆),我们将所有元素入队,每次取前两个捆,捆完后入队即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <c…